Question: Evaluate the triple integral. $ \int_{-2}^0 \int_1^4 \int_{-x}^0 4z - 2y \, dz \, dx \, dy =$
We can evaluate triple integrals by repeated integration: $ \int_{a_0}^{a_1} \int_{b_0}^{b_1} \int_{c_0}^{c_1} f(x, y, z) \, dx \, dy \, dz = \int_{a_0}^{a_1} \left( \int_{b_0}^{b_1} \left[ \int_{c_0}^{c_1} f(x, y, z) \, dx \right] dy \right) dz$ The first layer: $\begin{aligned} &\int_{-2}^0 \int_1^4 \int_{-x}^0 4z - 2y \, dz \, dx \, dy \\ \\ &= \int_{-2}^0 \int_1^4 \left[ 2z^2 - 2zy \right]_{-x}^0 dx \, dy \\ \\ &= \int_{-2}^0 \int_1^4 -2x^2 - 2xy \, dx \, dy \end{aligned}$ The second layer: $\begin{aligned} &\int_{-2}^0 \int_1^4 -2x^2 - 2xy \, dx \, dy \\ \\ &= \int_{-2}^0 \left[ \dfrac{-2x^3}{3} - x^2y \right]_1^4 dy \\ \\ &= \int_{-2}^0 \dfrac{-128}{3} - 16y - \left( \dfrac{-2}{3} - y \right) dy \\ \\ &= \int_{-2}^0 -15y - 42 \, dy \end{aligned}$ The third layer: $\begin{aligned} \int_{-2}^0 -15y - 42 \, dy &= \left[ \dfrac{-15y^2}{2} - 42y \right]_{-2}^0 \\ \\ &= -\left( \dfrac{-60}{2} + 84 \right) \\ \\ &= -54 \end{aligned}$ In conclusion: $ \int_{-2}^0 \int_1^4 \int_{-x}^0 4z - 2y \, dz \, dx \, dy = -54$